Question

A boy is sitting on a swing and blowing a whistle at a frequency of $1000\text{Hz}$. The swing is moving through an angle of ${30}^{\circ }$ from vertical. The boy is at $2\text{m}$ from the point of support of the swing and a girl stands in front of the swing. Then the maximum frequency heard by the girl is: (Given: velocity of sound in air = $330\text{m/s}$)


[Neglect, velocity of wind and take $g=10{\text{m/s}}^2$]

• A. $1000\text{Hz}$
• B. $1001\text{Hz}$
• C. $1007\text{Hz}$
• D. $1011\text{Hz}$

Answer 1

The correct option is C $1007\text{Hz}$

From equation of Doppler effect,
Apparant frequency $f_0=\frac{v\pm v_o}{v\mp v_s}$

From the data given in the question, we can say that,

Apparent frequency, $f^{\prime }=f_0\left(\frac{v}{v\pm vs}\right)$
[since the observer is at rest]

Maximum frequency is heard when the swing is at mean position (max. velocity) and approaching the observer.

When the source is at the mean position moving away, we hear minimum frequency.

So, we can write that,

$f_{max}=f_0\left(\frac{v}{v-v_{max}}\right)$

Applying Work-Energy theorem for the swing,

$\frac{1}{2}m{v}_{max}^2=mgh\Rightarrow v_{max}^2=2gh$

$\Rightarrow v_{max}^2=2g\times 2(1-\cos 30{}^{\circ })$

$\Rightarrow v_{max}^2=5.36\Rightarrow v_{max}=2.315\text{m/s}$

From the data given in the question,

$f_0=1000\text{Hz},v=330\text{m/s},v_{max}=2.32\text{m/s}$

Hence, $f_{max}=1000\times \frac{330}{(330-2.315)}$

$f_{max}=1000\times \frac{330}{327.675}\approx 1007\text{Hz}$